f(x)=x-2lnx
[tex]f'(x)=1-\frac{2}{x}=\frac{x-2}{x}[/tex]
(vezi tabelul de derivare din atasament)
Monotonia functiei f
Calculam f'(x)=0
[tex]\frac{x-2}{x}=0[/tex]
x-2=0
x=2
Facem tabel semn
x 1 2 +∞
f'(x) - - - - - 0 + + + + + +
f(x) ↓ f(2) ↑
2-2ln2
f(2)=2-2ln2 -punctul de extrem local
Convexitatea
Calculam f''(x)
[tex]f''(x)=(1-\frac{2}{x})'=0-(-\frac{2}{x^2})=\frac{2}{x^2}[/tex]
Observam ca f''(x)>0⇒ f(x) este convexa pe [1,+∞)
f(e)=e-2lne=e-2
[tex]\lim_{x \to e} \frac{f(x)-f(e)}{x^2-e^2}=\frac{0}{0}[/tex]
Aplicam L'Hospital (derivam numarator, derivam numitor)
[tex]\lim_{x \to e} \frac{f'(x)-0}{2x}= \lim_{x \to e} \frac{\frac{x-2}{x} }{2x} = \lim_{x \to e} \frac{x-2}{2x^2}=\frac{e-2}{2e^2}[/tex]
sau
[tex]\lim_{x \to e} \frac{f(x)-f(e)}{(x-e)(x+e)} = f'(e)\cdot \lim_{x \to e} \frac{1}{x+e} =\frac{e-2}{e}\cdot \frac{1}{2e}=\frac{e-2}{2e^2}[/tex]
Ne folosim de faptul ca:
[tex]\lim_{x \to e} \frac{f(x)-f(e)}{x-e}=f'(e)[/tex]
Mai multe despre monotonia unei functii gasesti aici: https://brainly.ro/tema/978074
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